How to return first async task to complete

Question

I've got several tasks that I'm running asynchronously. Depending on the input, one or more might run long, but only one of the tasks will return the :success message.

slowtask = Task.async(slow())
fasttask = Task.async(fast())

How can I capture the first of the two tasks above to complete, without having to wait for the other? I've tried Task.find/2, but because its implemented with enum, it seems to wait for all exit signals before finding a ref/message. My other thought was to poll this in Stream.cycle, ignoring still alive tasks and catching one that has exited. It seems un elixir like to poll in this way though.

Answer 1

There is no easy way to do this on Elixir yet. Your best option is, if you are only waiting for those messages in a given process, is something like this:

  defmodule TaskFinder do
    def run do
      task1 = Task.async fn -> :timer.sleep(1000); 1 end
      task2 = Task.async fn -> :timer.sleep(5000); 2 end
      await [task1, task2]
    end

    # Be careful, this will receive all messages sent
    # to this process. It will return the first task
    # reply and the list of tasks that came second.
    def await(tasks) do
      receive do
        message ->
          case Task.find(tasks, message) do
            {reply, task} ->
              {reply, List.delete(tasks, task)}
            nil ->
              await(tasks)
          end
      end
    end
  end

  IO.inspect TaskFinder.run

Note you could also use this pattern to spawn tasks in a GenServer and use Task.find/2 to find the matching ones. I have also added this example to Elixir docs.